Integration of Trigonometric Functions
Download as PPTX, PDF8 likes3,513 views
This document discusses integration using algebraic substitution and trigonometric substitution. It provides examples of integrating trigonometric functions and expressions involving radicals using these substitution techniques. Key formulas for integrating trigonometric functions are presented. Examples show how to transform integrals involving powers of trigonometric functions into integrable forms by using trigonometric identities.
Ad
More Related Content
What's hot (20)
Viewers also liked (20)
![Jacob riis[1] (1) lol](https://cdn.slidesharecdn.com/ss_thumbnails/jacobriis11lol-140525002816-phpapp01-thumbnail.jpg?width=560&fit=bounds)
Ad
Similar to Integration of Trigonometric Functions (20)
Ad
More from Raymundo Raymund (6)
Recently uploaded (20)
Integration of Trigonometric Functions
- 1. Integration of Trigonometric Functions RAYMUND T. DE LA CRUZ MAEd-mathematics
- 2. Integration by Algebraic Substitution>>> By noting the formulas for differentiating the six trigonometric functions, and the antiderivatives that are found using them, we have the following integration formulas: sin 𝑢 𝑑𝑢 = − cos 𝑢 + 𝐶 cos 𝑢 𝑑𝑢 = sin 𝑢 + 𝐶 sec2 𝑢 𝑑𝑢 = tan 𝑢 + 𝐶 csc2 𝑢 𝑑𝑢 = − cot 𝑢 + 𝐶 sec 𝑢 tan 𝑢 𝑑𝑢 = sec 𝑢 + 𝐶 csc 𝑢 cot 𝑢 𝑑𝑢 = − csc 𝑢 + 𝐶
- 3. Integration by Algebraic Substitution>>> Example1 Integrate: 𝑥 𝑠𝑒𝑐2 𝑥2 𝑑𝑥. With 𝑢 = 𝑥2 , 𝑑𝑢 = 2𝑥 𝑑𝑥, we have 𝑥 𝑠𝑒𝑐2 𝑥2 𝑑𝑥 = 1 2 (𝑠𝑒𝑐2 𝑥2 (2𝑥 𝑑𝑥) = 1 2 tan 𝑥2 + 𝐶 Example2 Integrate: tan 2𝑥 cos 2𝑥 𝑑𝑥. By using the basic identity sec 𝜃 = 1 cos 𝜃 , we can transform this integral into the form sec2𝑥 tan2𝑥 𝑑𝑥. In this form, 𝑢 = 2𝑥, 𝑑𝑢 = 2 𝑑𝑥. Therefore, tan 2𝑥 cos 2𝑥 𝑑𝑥 = sec2𝑥 tan2𝑥 𝑑𝑥 = 1 2 sec2𝑥 tan2𝑥 2 𝑑𝑥 = 1 2 sec2𝑥 + 𝐶
- 4. Integration by Algebraic Substitution>>> We also have the following integration formulas: tan 𝑢 𝑑𝑢 = − 𝑙𝑛 cos 𝑢 + 𝐶 cot 𝑢 𝑑𝑢 = ln sin 𝑢 + 𝐶 sec 𝑢 𝑑𝑢 = ln sec 𝑢 + tan 𝑢 + 𝐶 csc 𝑢 𝑑𝑢 = ln csc 𝑢 − cot 𝑢 + 𝐶 Example3 Integrate: tan 4𝜃 𝑑𝜃. Noting that 𝑢 = 4𝜃, 𝑑𝑢 = 4𝑑𝜃, we have tan 4𝜃 𝑑𝜃 = 1 4 tan 4𝜃(4𝑑𝜃) = − 1 4 ln cos 4𝜃 + 𝐶
- 5. Integration by Algebraic Substitution>>> By use of trigonometric relations, it is possible to transform many integrals involving powers of the trigonometric functions into integrable form. We now show the relationships that are useful for these integrals. 𝑐𝑜𝑠2 𝑥 + 𝑠𝑖𝑛2 𝑥 = 1 1 + 𝑡𝑎𝑛2 𝑥 = 𝑠𝑒𝑐2 𝑥 1 + 𝑐𝑜𝑡2 𝑥 = 𝑐𝑠𝑐2 𝑥 2𝑐𝑜𝑠2 𝑥 = 1 + cos 2𝑥 2𝑠𝑖𝑛2 𝑥 = 1 − cos 2𝑥 Example1 Integration with odd power of sin u Integrate: 𝑠𝑖𝑛3 𝑥 𝑐𝑜𝑠2 𝑥 𝑑𝑥. Because 𝑠𝑖𝑛3 𝑥 = 𝑠𝑖𝑛2 𝑥 sin 𝑥 = 1 − 𝑐𝑜𝑠2 𝑥 sin 𝑥, we can write this integral with powers of cos x along with – sin x, which is the necessary 𝑑𝑢 for this integral. Therefore, 𝑠𝑖𝑛3 𝑥 𝑐𝑜𝑠2 𝑥 𝑑𝑥 = (1 − 𝑐𝑜𝑠2 𝑥)(sin 𝑥) 𝑐𝑜𝑠2 𝑥 𝑑𝑥 = (𝑐𝑜𝑠2 𝑥 − 𝑐𝑜𝑠4 𝑥)(sin 𝑥 𝑑𝑥) =− 𝑐𝑜𝑠2 𝑥(− sin 𝑥 𝑑𝑥) + 𝑐𝑜𝑠4 𝑥(− sin 𝑥 𝑑𝑥) = − 1 3 𝑐𝑜𝑠3 𝑥 + 1 5 𝑐𝑜𝑠5 𝑥 + 𝐶
- 6. Integration by Algebraic Substitution>>> Example2 Integration with odd power of cos u Integrate: 𝑐𝑜𝑠5 2𝑥 𝑑𝑥. Because 𝑐𝑜𝑠5 2𝑥 = 𝑐𝑜𝑠4 2𝑥 cos 2𝑥 = (1 − 𝑠𝑖𝑛2 2𝑥)2 cos 2𝑥, it is possible to write this integral with powers of sin 2x along with cos 2x dx. Thus, with the introduction of a factor of 2, (cos 2x)(2dx) is the necessary 𝑑𝑢 for this integral. Thus, 𝑐𝑜𝑠5 2𝑥 𝑑𝑥 = (1 − 𝑠𝑖𝑛2 2𝑥)2 cos 2𝑥 𝑑𝑥 = 1 − 2𝑠𝑖𝑛2 2𝑥 + 𝑠𝑖𝑛4 2𝑥 cos 2𝑥 𝑑𝑥 = cos 2𝑥 𝑑𝑥 − 2𝑠𝑖𝑛2 2𝑥 cos 2𝑥 𝑑𝑥 + 𝑠𝑖𝑛4 2𝑥 cos 2𝑥 𝑑𝑥 = 1 2 cos 2𝑥 2𝑑𝑥 − 𝑠𝑖𝑛2 2𝑥 (2cos 2𝑥 𝑑𝑥) + 1 2 𝑠𝑖𝑛4 2𝑥 (2cos 2𝑥 𝑑𝑥) = 1 2 𝑠𝑖𝑛 2𝑥 − 1 3 𝑠𝑖𝑛3 2𝑥 + 1 10 𝑠𝑖𝑛5 2𝑥 + 𝐶
- 7. Integration by Algebraic Substitution>>> Example4 Integration with power of sec u Integrate: 𝑠𝑒𝑐3 𝑡 tan 𝑡 𝑑𝑡. By writing 𝑠𝑒𝑐3 𝑡 tan 𝑡 as 𝑠𝑒𝑐2 𝑡 (sec 𝑡 tan 𝑡), we can use the sec 𝑡 tan 𝑡 𝑑𝑡 as the 𝑑𝑢 of the integral. Thus, 𝑠𝑒𝑐3 𝑡 tan 𝑡 𝑑𝑡 = 𝑠𝑒𝑐2 𝑡 (sec 𝑡 tan 𝑡 𝑑𝑡) = 1 3 𝑠𝑒𝑐3 𝑡 + 𝐶 Example3 Integration with even power of sin u Integrate: 𝑠𝑖𝑛2 2𝑥 𝑑𝑥. Since 𝑠𝑖𝑛2 2𝑥 = 1 2 (1 − cos 4𝑥), this integral can be transformed into a form that can be integrated. Therefore we write 𝑠𝑖𝑛2 2𝑥 𝑑𝑥 = 1 2 (1 − cos 4𝑥) 𝑑𝑥 = 1 2 𝑑𝑥 − 1 8 cos 4𝑥(4𝑑𝑥) = 𝑥 2 − 1 8 sin 4𝑥 + 𝐶
- 8. Integration by Algebraic Substitution>>> Example5 Integration with power of tan u Integrate: 𝑡𝑎𝑛5 𝑥 𝑑𝑥. Because 𝑡𝑎𝑛5 𝑥 = 𝑡𝑎𝑛3 𝑥 𝑡𝑎𝑛2 𝑥 = 𝑡𝑎𝑛3 𝑠𝑒𝑐2 − 1 , we can write this integral with powers of tan 𝑥 along with 𝑠𝑒𝑐2 𝑥 𝑑𝑥. Thus, 𝑠𝑒𝑐2 𝑥 𝑑𝑥 becomes the necessary 𝑑𝑢 of the integral. It is necessary to replace 𝑡𝑎𝑛2 𝑥 with 𝑠𝑒𝑐2 𝑥 − 1 twice during the integration. Therefore, 𝑡𝑎𝑛5 𝑥 𝑑𝑥 = 𝑡𝑎𝑛3 𝑠𝑒𝑐2 − 1 𝑑𝑥 = 𝑡𝑎𝑛3 𝑥 (𝑠𝑒𝑐2 𝑥 𝑑𝑥) − 𝑡𝑎𝑛3 𝑥 𝑑𝑥 = 1 4 𝑡𝑎𝑛4 𝑥 − tan 𝑥 𝑠𝑒𝑐2 𝑥 − 1 𝑑𝑥 = 1 4 𝑡𝑎𝑛4 𝑥 − tan 𝑥(𝑠𝑒𝑐2 𝑥 𝑑𝑥) + tan 𝑥 𝑑𝑥 = 1 4 𝑡𝑎𝑛4 𝑥 − 1 2 𝑡𝑎𝑛2 𝑥 − ln cos 𝑥 + 𝐶 P r a c t ic e Ex e r c is e Integrate: 1. sin5𝑥 𝑑𝑥 4. 𝑠𝑖𝑛3 𝑥 𝑑𝑥 2. 6𝑐𝑠𝑐2 3𝑥 𝑑𝑥 5. 𝑠𝑒𝑐4 𝑥 𝑑𝑥 3. 4𝑥 cot 𝑥2 𝑑𝑥
- 9. Integration by Trigonometric Substitution>>> Substitutions based on trigonometric relations are particularly useful for integrating expressions involving radicals. Example1 For 𝑎2 − 𝑥2, let 𝑥 = 𝑎 sin 𝜃 Integrate: 𝑑𝑥 𝑥2 1−𝑥2 . If we let 𝑥 = sin 𝜃, then 1 − 𝑠𝑖𝑛2 𝜃 = cos 𝜃, and the integral can be transformed into a trigonometric integral. Carefully replacing all factors of the integral with expressions in terms of 𝜃, we have 𝑥 = sin 𝜃, 1 − 𝑥2 = cos 𝜃, and 𝑑𝑥 = cos 𝜃 𝑑𝜃. Therefore, 𝑑𝑥 𝑥2 1−𝑥2 = cos 𝜃 𝑑𝜃 𝑠𝑖𝑛 2 𝜃 1−𝑠𝑖𝑛 2 𝜃 = 𝑐𝑜𝑠𝜃 𝑑𝜃 𝑠𝑖𝑛 2 𝜃 cos 𝜃 = 𝑐𝑠𝑐2 𝜃 𝑑𝜃 = − cot 𝜃 + 𝐶 We have now performed the integration, but the answer we now have is in terms of 𝜃, and we must express the result in terms of x. Making a triangle with an angle 𝜃 such that sin 𝜃 = 𝑥 1 , we may express any of the trigonometric functions in terms of x. Thus, cot 𝜃 = 1−𝑥2 𝑥 Therefore, the result of the integration becomes 𝑑𝑥 𝑥2 1−𝑥2 = − cot 𝜃 + 𝐶 = − 1−𝑥2 𝑥 + 𝐶
- 10. Integration by Trigonometric Substitution>>> Example2 For 𝑎2 + 𝑥2, let 𝑥 = 𝑎 tan 𝜃 Integrate: 𝑑𝑥 𝑥2+4 . If we let 𝑥 = 2 tan 𝜃, the radical in this integral becomes 𝑥2 + 4 = 4 𝑡𝑎𝑛2 𝜃 + 4 = 2 𝑡𝑎𝑛2 𝜃 + 1 = 2 𝑠𝑒𝑐2 𝜃 = 2 sec 𝜃 Therefore, with 𝑥 = 2 tan 𝜃 and 𝑑𝑥 = 2 𝑠𝑒𝑐2 𝜃 𝑑𝜃, we have 𝑑𝑥 𝑥2+4 = 2 𝑠𝑒𝑐 2 𝜃 𝑑𝜃 4 𝑡𝑎𝑛 2 𝜃+4 = 2 𝑠𝑒𝑐 2 𝜃 𝑑𝜃 2 sec 𝜃 = sec 𝜃 𝑑𝜃 = ln sec 𝜃 + tan 𝜃 + 𝐶 = ln 𝑥2+4 2 + 𝑥 2 + 𝐶 = ln 𝑥2+4+𝑥 2 + 𝐶 This answer is acceptable, but by using the properties of logarithms, we have ln 𝑥2+4+𝑥 2 + 𝐶 = ln 𝑥2 + 4 + 𝑥 + (𝐶 − ln 2) = ln 𝑥2 + 4 + 𝑥 + 𝐶′
- 11. Integration by Trigonometric Substitution>>> Example3 For 𝑥2 − 𝑎2, let 𝑥 = 𝑎 sec 𝜃 Integrate: 2 𝑑𝑥 𝑥 𝑥2−9 . If we let 𝑥 = 3 sec 𝜃, the radical in this integral becomes 𝑥2 − 9 = 9 𝑠𝑒𝑐2 𝜃 − 9 = 3 𝑠𝑒𝑐2 𝜃 − 1 = 3 𝑡𝑎𝑛2 𝜃 = 3 tan 𝜃 Therefore, with 𝑥 = 3 sec 𝜃 tan 𝜃 𝑑𝜃, we have 2 𝑑𝑥 𝑥 𝑥2−9 = 2 3 sec 𝜃 tan 𝜃 𝑑𝜃 3 sec 𝜃 9𝑠𝑒𝑐 2 𝜃−9 = 2 tan 𝜃 𝑑𝜃 3 tan 𝜃 = 2 3 𝑑𝜃 = 2 3 𝜃 + 𝐶 = 2 3 𝑠𝑒𝑐−1 𝑥 3 + 𝐶 P r a c t ic e Ex e r c is e Integrate: 1. 1−𝑥2 𝑥2 𝑑𝑥 4. 6𝑑𝑧 𝑧2 𝑧2+9 2. 2𝑑𝑥 𝑥2−36 5. 6𝑑𝑥 𝑥 4−𝑥2 3. 𝑥2−25 𝑥 𝑑𝑥